If a,b,c and d are positive real numbers such that logab=8/9, logbc=-3/4, logcd=2, find the value of logd(abc)I really need help on this

We can expand the logarithm of a product as a sum of logarithms:[tex]\log_dabc=\log_da+\log_db+\log_dc[/tex]Then using the change of base formula, we can derive the relationship[tex]\log_xy=\dfrac{\ln y}{\ln x}=\dfrac1{\frac{\ln x}{\ln y}}=\dfrac1{\log_yx}[/tex]This immediately tells us that[tex]\log_dc=\dfrac1{\log_cd}=\dfrac12[/tex]Notice that none of [tex]a,b,c,d[/tex] can be equal to 1. This is because[tex]\log_1x=y\implies1^{\log_1x}=1^y\implies x=1[/tex]for any choice of [tex]y[/tex]. This means we can safely do the following without worrying about division by 0.[tex]\log_db=\dfrac{\ln b}{\ln d}=\dfrac{\frac{\ln b}{\ln c}}{\frac{\ln d}{\ln c}}=\dfrac{\log_cb}{\log_cd}=\dfrac1{\log_bc\log_cd}[/tex]so that[tex]\log_db=\dfrac1{-\frac34\cdot2}=-\dfrac23[/tex]Similarly,[tex]\log_da=\dfrac{\ln a}{\ln d}=\dfrac{\frac{\ln a}{\ln b}}{\frac{\ln d}{\ln b}}=\dfrac{\log_ba}{\log_bd}=\dfrac{\log_db}{\log_ab}[/tex]so that[tex]\log_da=\dfrac{-\frac23}{\frac89}=-\dfrac34[/tex]So we end up with[tex]\log_dabc=-\dfrac34-\dfrac23+\dfrac12=-\dfrac{11}{12}[/tex]###Another way to do this:[tex]\log_ab=\dfrac89\implies a^{8/9}=b\implies a=b^{9/8}[/tex][tex]\log_bc=-\dfrac34\implies b^{-3/4}=c\implies b=c^{-4/3}[/tex][tex]\log_cd=2\implies c^2=d\implies\log_dc^2=1\implies\log_dc=\dfrac12[/tex]Then[tex]abc=(c^{-4/3})^{9/8}c^{-4/3}c=c^{-11/6}[/tex]So we have[tex]\log_dabc=\log_dc^{-11/6}=-\dfrac{11}6\log_dc=-\dfrac{11}6\cdot\dfrac12=-\dfrac{11}{12}[/tex]